5 g of steam at 100 C is passed into 100 g of water at 20 C (L v = 540 cal/g, c = 1 cal/g C). The final temperature is about:

⚛️ Physics · Thermal Properties of Matter · NEET & JEE

5 g of steam at 100°C is passed into 100 g of water at 20°C (L<sub>v</sub> = 540 cal/g, c = 1 cal/g°C). The final temperature is about:

Answer: 49.5 C.

A 55°C
B 49.5°C
C 45°C
D 40°C

Correct answer: B. 49.5°C

Explanation: Heat from steam: 5×540 + 5(100-T); heat to water: 100(T-20). Solving gives T ≈ 49.5°C.

During a phase change (melting or boiling), temperature stays constant while heat is absorbed entirely as latent heat (Q=mL); temperature only rises again once the substance is fully in its new phase.

Concept context

Temperature scales, thermal expansion, calorimetry, and the three modes of heat transfer including radiation laws.

Read the full Thermal Properties of Matter notes →

5 g of steam at 100&deg;C is passed into 100 g of water at 20&deg;C (L<sub>v</sub> = 540 cal/g, c = 1 cal/g&deg;C). The final temperature is about:

Concept context

More Thermal Properties of Matter questions

5 g of steam at 100°C is passed into 100 g of water at 20°C (L<sub>v</sub> = 540 cal/g, c = 1 cal/g°C). The final temperature is about: